So now that we've explored the many techniques associated with antidifferentiating a function or group or functions, we can now apply them calculate any complex definite integrals we might encounter. But how would you calculate a definite integral which we don't necessarily know the antiderivative for? Such as:
If you stop and think for a second, there is actually another way to solve such a definite integral without applying the techniques of antidifferentiation (since determining the antiderivative of the above function has so far proved quite impossible). As we were first introduced to the concept of the integral, we focused mainly on the geometric interpretation of various functions. This can easily be applied to the above function, by using Riemannian sums on the above equation, a definite value can in fact be calculated. The main concern is determining just how close we can get to this value in the simplest manner or using the simplest techniques.
Digress from the above analysis on non-antidifferentiable functions and consider the following functions.
Now though these functions aren't difficult to antidifferentiate at all, they will be quite an aid for analyzing the various methods we have been introduced to when calculating definite integrals. The first of which, concerns the left and right-hand sums of the given function.
***Before determining the left and right hand sums for each function, recall the significance of monotonicity. What is meant by a monotonic increasing function? What is meant by a monotonic decreasing function? How do they affect approximations?
If you can't quite remember what a monotonic increasing/decreasing function represents, the underlying or basic idea of such a concept is not too difficult to grasp, though it has been a while since this concept has been applied in any way.
-- If a function f on the interval [a,b] is increasing throughout the interval, then that function is said to be monotonic increasing.
-- If a function f on the interval [a,b] is decreasing throughout the interval, then that function is said to be monotonic decreasing.
You may also take note of the fact that if a function f is increasing on a given interval [a,b] (monotonic increasing), then the function's derivative f' is > 0 throughout that interval. Likewise for a monotonic decreasing function where it's derivative is negative throughout.
Left and Right hand sums
Now back to the main concept at hand; to left and right hand sums. If you are not familiar with the process of calculating left and right hand sums or you have forgotten but have the RSUM / RIESUM program on your graphing calculator, here's a breakdown on how to use the program.
- Hit the [y=] button, then insert the given function in the Y1 slot.
- Hit [prgm] and then choose the RSUM / RIESUM program.
- It will prompt you to insert a "LEFT" value, this is the lower limit of the integral.
- It will then prompt you to insert a "RIGHT" value, which is now the upper limit of the integral.
- Next will be a prompt for you to insert a "X CHOICE" value, which indicates the type of sum desired. For a left sum, X CHOICE = 0, for a midpoint sum X CHOICE = 0.5, for a right sum X CHOICE = 1. Try not to forget this.
- It will finally ask for a "NUMBER" value, this is the number of desired sub intervals.
- Hit enter, and it will output the value of the calculated sum, whether it be left, right or mid.
The following graphs show how the left and right hand sums compare to the actual graphs and also to each other.
Do you notice any patterns lying within the above graphs? One important factor to notice is that one function is monotonic increasing, while the other is monotonic decreasing. This does in fact play an important role in the accuracy and orientation of a particular left or right hand sum with regard to the main function. Here's a breakdown on this connection.For a monotonic increasing function:
-- The left hand sum will always be an underestimate. This means that using a left hand sum will always be slightly less than the actual value of the integral over that interval.
-- The right hand sum will always be an overestimate. This means that using a right hand sum will always be slightly greater than the actual value of the integral over that interval.
For a monotonic decreasing function:
-- The left hand sum will always be an overestimate.
-- The right hand sum will always be an underestimate.
At this point you might be asking yourself which is better, a left hand sum or a right hand sum. But do these answers have any correlation with the tonicity of the function in question?
Well if you analyze the left and right hand sums for both monotonic increasing and monotonic decreasing functions above, they appear as though both sums produce approximately the same amount of error. But if you take into account the fact that a trapezoid sum (discussed further below) is the average between these two sums (left and right), then you can actually determine which sum truly retains a greater error. This can be achieved by analyzing where a trapezoid sum is an under or an overestimate, now since this sum is the average between a left and a right sum then this under or overestimate in the trapezoid sum directly reflects which sum has a greater error for a given function. This basically means that if the error of the trapezoid sum is known to be an underestimate, then whichever sum yielded an underestimate therefore had the greater error, and likewise for a trapezoid sum that is an overestimate. Now, I don't think that it's important to memorize this fact, but just know the basic concepts of analyzing the geometric aspect of functions and you should be fine.
One might argue that this could be compensated for by introduced a greater amount of sub intervals, just as Riemann had showed the world. But how could you compensate for such a deficit with a fixed number of sub intervals? Basically using varied width for each sub interval could work. Since Riemann had basically surmised that it does not matter how wide each sub interval is, just as long as you go to infinity, that's what matters. So now that the idea of possibly using varied sub interval widths has quickly implanted itself into your mind and has just as quickly departed, beware that it's possible that a question involving such varied sub intervals might pop up on the AP exam. Remember, using Riemann sums and determining an integral by hand relies solely on your interpretation of the geometry of the function in question. Remember that.
Trapezoid Sums
Recall another sum that was quite important in calculating definite integrals was the trapezoid sum. We were both informally introduced to this concept as being the mean of the left and right hand sums for the function and also formally introduced to the fact that averaging the left and right hand sums creates a series of trapezoid in order to approximate the area under the curve. This technique is sufficiently more approximate than just a left or a right hand sum, but is it right on or is there some inaccuracy associated with a trapezoid sum? Of course there is inaccuracy when using any type of sum that we have encountered so far, but which sum has the greatest accuracy?
Here is an example depicting the idea behind a trapezoid sum:
Now we must find the trapezoid sums for the above functions in order to aptly contrast each process.
Notice by the above graphs that for an monotonic increasing function, the trapezoid sum is a slight overestimate, and that for a monotonic decreasing function the trapezoid sum is a slight underestimate. By looking at the graphs, it would appear that the trapezoid sum is overall the best sum by far. But what about the final type of sum that we have encountered throughout our study of integration. This sum is of course the midpoint sum.
Midpoint Sums
Do you remember what the purpose or the main idea behind a midpoint sum is? Basically, instead of using rectangles that use the left or right hand values of each sub interval as their height, the rectangles take upon themselves the value of the midpoint of each sub interval as their height. Below are two example of how a midpoint sum works, where the second example uses a tangent line instead of a straight line.
Also, as with the above summing techniques, applying them to the aforementioned functions is required to better receive a deeper understanding of how these processes compare to each other. 
Now take note that inaccuracy attained by using a midpoint sum apparently rivals that of a trapezoid sum. This is true since a midpoint sum for a monotonic increasing function is always an underestimate, conversely a midpoint sum for a monotonic decreasing function yields an overestimate. Now we've covered the main sum techniques encountered so far this year, it's time to go a step further.
The Best Approximation?
So which technique appears to be the best? So far, it's either going to be the midpoint sum technique or the trapezoid technique. A possible way to determine which is truly more precise, apply them both to an integral with a known value. Setting up a table for different amounts of sub intervals and also with positions to record the given error of each respective summing technique will be essential in conducting this analysis. For the integral below, the value is known to be exactly 1/4, so by subtracting the received value from each calculation from 1/4 you can acquire the amount of error accompanying the calculation. Take note that it's now possible to receive both positive and negative errors, but let positive error values represent overestimates and negative error values represent underestimates.
Under close inspection of the error of using each sum technique, it appears as though a midpoint sum is twice as accurate as a trapezoid sum. So now we know how we can find the best approximation possible when antidifferentiation is impossible at our level. Since a midpoint sum yields half the error, and is therefore twice as better, is appears as though midpoint sums are the ideal candidates for numerical integration. Though there must be better techniques, such as a technique that will NOT be on the AP exam, but will be prominent in higher level calculus classes. This technique is known as the Simpson sum.The idea behind a Simpson sum is simple. I must note, however, that it must be understood that the minimum amount of points required to configure a straight line is two, while the minimum amount of points required to configure a parabolic curve is three, this being an important idea behind a Simpson sum. Now, since using a midpoint sum yields such a great accuracy using a line that connects two points, what if you were to use curves with three points to attempt to fit the curve? Basically, instead of having a straight line connecting two points within a sub interval, you can use three points connecting two lines within that sub interval. Now since a trapezoid sum's inaccuracy is always the opposite the inaccuracy of the midpoint sum, but twice the value, if you add the trapezoid sum to twice the midpoint sum and divide that value by three, the error would approach 0 significantly faster than with a trapezoid or midpoint sum thus producing remarkable accuracy.
Now, formal introductions to the idea of trapezoid and midpoint sum error calculations are necessary. These error calculations are as follows:
______________________________________________________________Wow, I'm finally finished scribing. I didn't think that this post would be so long, and I really wanted to complete this scribe using notebook software or something else, but I was having a great amount of tribulation when trying to export, publish, upload, etc. Well anyway, this is pretty much complete now, all that's left is to state who the next scribe will be! Well goodnight and goodbye everyone! Have a great weekend.
Ah yes, the next scribe will be :
GreyM
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